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↑↑ Dacu:
How does from the fact that , are natural numbers follows that , are natural numbers?
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Dear colleagues, I certainly do not want to interfere with this discussion, however, Dacu - if you are sure you have an alternative proof of FLT, I strongly recommend to submit it into a suitable journal, in this case probably to the Annals of Mathematics.
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It can be useful excercise to find weak spots in presented proofs, so we can continue this discussion.
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Hello all,
Dear users of the forum, I will soon show that from variant "2) and where with where and ". results .
For personal reasons I have to postpone the next post. Thank you very much for your understanding!
All the best,
Dacu
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↑ Dacu:
Firstly you have to proove that and that .
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check_drummer napsal(a):
↑ Dacu:
Firstly you have to proove that and that .
Hello,
Because in 5) we have and , then it follows that the equation is a Diophantine equation with rational coefficients.
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Example of Diophantine equation with rational coefficients:
Solve the Diophantine equation in the set of integer numbers .
Solution:
and where and .
I am waiting for the following questions ...
Thank you very much!
All the best,
Dacu
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↑ Dacu:
Hi,
1) why do both inequalities hold?
and why it is important that they hold?
2) Why it is important that we got Diophantine equation?
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check_drummer napsal(a):
↑ Dacu:
Hi,
1) why do both inequalities hold?
and why it is important that they hold?
2) Why it is important that we got Diophantine equation?
Hello,
1) Correction:
and
2) For demonstration .... and in fact is a Diophantine equation ...
All the best,
Dacu
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↑ Dacu:
So why does the inequalities with absolute values hold?
Edit: I see inequality "<1" - because we are subtracting numbers that are in interval (0;1), but I don't see inequality "<0" ie why those numbers are not equal.
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check_drummer napsal(a):
↑ Dacu:
So why does the inequalities with absolute values hold?
Edit: I see inequality "<1" - because we are subtracting numbers that are in interval (0;1), but I don't see inequality "<0" ie why those numbers are not equal.
Hello,
I repeat:
"From the following possibilities result:
1) and .
2) and where with where and .
Which of the two possibilities can exist, 1) or 2)?"
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What inequality are you talking about, what equality of numbers are you talking about?Thank you very much!
All the best,
Dacu
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↑ Dacu:
So - you dont know the answer and this is the question for the whole forum?
Why in 2) this must hold: , ?
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check_drummer napsal(a):
↑ Dacu:
So - you dont know the answer and this is the question for the whole forum?
Why in 2) this must hold: , ?
Hello,
I reformulate variant 2):
2) and where and .
Let and two irreducible fractions , then there are the following possibilities:
a) , , and where with .
b) , , and where where .
Because and , , must have no common divisors, then it is necessary that and so which contradicts .
All the best,
Dacu
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↑ Dacu:
So you are trying to prove that variant 2) cannot hold, right? So I have two questions:
I) How do you know that is whole number and not rational number? (From what fact daes it follow?)
II) Why other variants except a) and b) cannot hold - ie that p,q have the same sign?
III) Why eg in a) equality holds?
Thank you.
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